Computing Volume Between a Surface and a Plane

Volume as Height Above a Rectangular Base

A double integral can measure more than the accumulated values of a function. When a function describes the vertical height of a surface above a rectangle, integrating that height gives the volume beneath the surface. To find the volume between a surface and a horizontal plane, the relevant height is the difference between the two vertical levels.

Let $R=[a,b]\times[c,d]$ be a rectangle in the $xy$-plane, let $z=f(x,y)$ be a continuous surface over $R$, and let $z=k$ be a horizontal plane. At each point $(x,y)$, the vertical distance between the graph and the plane is $|f(x,y)-k|$. Thus the volume is found by integrating this nonnegative height over the rectangle.

Definition: The volume between $z=f(x,y)$ and the horizontal plane $z=k$, above a rectangle $R$, is $$ V=\iint_R |f(x,y)-k|\,dA. $$ If $f(x,y)\ge k$ throughout $R$, this becomes $\iint_R(f-k)\,dA$. If $f(x,y)\le k$ throughout $R$, it becomes $\iint_R(k-f)\,dA$.

Here, $dA$ represents area in the base plane. Because the domain is a rectangle and the height function is continuous, the integral can be evaluated as either iterated integral: $\int_a^b\int_c^d |f(x,y)-k|\,dy\,dx$, or $\int_c^d\int_a^b |f(x,y)-k|\,dx\,dy$. The order may be chosen for convenience, provided the bounds describe the same rectangle.

Why Integrating the Height Gives Volume

First suppose the surface is at or above the plane everywhere on $R$, so $f(x,y)-k\ge 0$. The solid between them consists of the points whose base coordinates lie in $R$ and whose heights lie between the plane and the graph: $\{(x,y,z):(x,y)\in R,\ k\le z\le f(x,y)\}$.

Divide the rectangle into small subrectangles. Over each small piece, the solid is approximately a column: its base area is the area of that piece, and its height is approximately $f(x_i,y_i)-k$ at a chosen point $(x_i,y_i)$ in the piece. The sum of these approximate column volumes is a rectangular Riemann sum for the height function. As the subdivisions become finer, continuity ensures that these sums approach the double integral. This explains why the height difference, rather than either surface height alone, must be integrated.

V=\lim \sum_{i,j}\bigl(f(x_{ij},y_{ij})-k\bigr)\,\Delta A_{ij} =\iint_R (f(x,y)-k)\,dA.

If the graph is below the plane, the same column argument gives height $k-f(x,y)$. If the graph crosses the plane, each column still has a nonnegative height, but that height is $|f(x,y)-k|$. In that case, integrating $f-k$ without accounting for its sign calculates a signed difference, not the volume.

A Reliable Setup

Before integrating, identify the domain, the two vertical levels, and which one is higher. The plane has constant height $k$; the graph may have a height that varies with both coordinates. The resulting vertical bounds are therefore allowed to depend on $x$ and $y$, even though the base domain is a rectangle.

Write the base rectangle.
Record the $x$- and $y$-intervals that specify $R$.

Compare the heights.
Determine whether $f-k$ is nonnegative everywhere, nonpositive everywhere, or changes sign.

Use the nonnegative vertical height.
Integrate $f-k$ if the surface is above the plane, $k-f$ if it is below, and $|f-k|$ if it crosses.

Evaluate and check.
Compute the iterated integral and confirm the result is nonnegative.

Worked Examples

Worked Example: A Surface Above the Plane

Find the volume between $z=2+x+y$ and the plane $z=2$ above $R=[0,2]\times[0,1]$.

The graph is $2+x+y$, so its height above the plane is $(2+x+y)-2=x+y$. On this rectangle, $x\ge0$ and $y\ge0$, so this height is nonnegative. The volume is

\begin{aligned} V &=\int_0^2\int_0^1 (x+y)\,dy\,dx\\ &=\int_0^2\left[xy+\frac{y^2}{2}\right]_{y=0}^{y=1}\,dx\\ &=\int_0^2\left(x+\frac12\right)\,dx\\ &=\left[\frac{x^2}{2}+\frac{x}{2}\right]_{x=0}^{x=2}\\ &=\left(2+1\right)-0=3. \end{aligned}

As a check, separate the two contributions: $\iint_R x\,dA=(\int_0^2x\,dx)(\int_0^1dy)=2$, and $\iint_R y\,dA=(\int_0^2dx)(\int_0^1y\,dy)=1$. Their sum is $2+1=3$, as computed. The volume is $3$ cubic units.

Worked Example: The Surface Crosses the Plane

Find the volume between $z=x$ and $z=1$ above $R=[0,2]\times[0,1]$.

The height difference $x-1$ is negative for $0\le x<1$ and positive for $1<x\le2$. The surface and plane meet at $x=1$. On the left part of the rectangle the plane is higher, and on the right part the graph is higher. The volume integral must use the positive height on each part:

\begin{aligned} V &=\int_0^1\int_0^1(1-x)\,dy\,dx +\int_1^2\int_0^1(x-1)\,dy\,dx\\ &=\int_0^1(1-x)\,dx+\int_1^2(x-1)\,dx\\ &=\left[x-\frac{x^2}{2}\right]_0^1 +\left[\frac{x^2}{2}-x\right]_1^2\\ &=\left(1-\frac12\right)-0 +\left((2-2)-\left(\frac12-1\right)\right)\\ &=\frac12+\frac12=1. \end{aligned}

The $y$-interval has width $1$, which is why integrating over $y$ leaves each height unchanged. Both pieces are nonnegative, and their sum is $1$ cubic unit. This also agrees with $\int_0^2|x-1|\,dx=1$, multiplied by the width $1$.

Worked Example: A Varying Surface Height

Find the volume between $z=4-x^2+y$ and the plane $z=1$ above $R=[-1,1]\times[0,2]$.

The height above the plane is $4-x^2+y-1=3-x^2+y$. Since $x^2\le1$ and $y\ge0$ on the rectangle, $3-x^2+y\ge2$; therefore, the graph is above the plane throughout. Integrate the height:

\begin{aligned} V &=\int_{-1}^1\int_0^2(3-x^2+y)\,dy\,dx\\ &=\int_{-1}^1\left[(3-x^2)y+\frac{y^2}{2}\right]_{y=0}^{y=2}\,dx\\ &=\int_{-1}^1(6-2x^2+2)\,dx\\ &=\left[8x-\frac{2x^3}{3}\right]_{-1}^1\\ &=\left(8-\frac23\right)-\left(-8+\frac23\right)\\ &=\frac{22}{3}-\left(-\frac{22}{3}\right)=\frac{44}{3}. \end{aligned}

For an independent check, integrate the three terms separately. The constant term contributes $3\cdot2\cdot2=12$; the term $-x^2$ contributes $-(\int_{-1}^1x^2\,dx)(\int_0^2dy)=-(2/3)(2)=-4/3$; and the term $y$ contributes $(\int_{-1}^1dx)(\int_0^2y\,dy)=2\cdot2=4$. Thus $12-4/3+4=44/3$, matching the result. The volume is $44/3$ cubic units.

A Common Mistake: Using Signed Height

When the surface crosses the plane, integrating $f-k$ over the whole rectangle can make positive and negative contributions cancel. For the preceding crossing example, that incorrect calculation would be

\int_0^2\int_0^1(x-1)\,dy\,dx =\int_0^2(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_0^2 =0.

Zero is not the volume: the solid has positive height on both sides of $x=1$. The calculation gives a signed integral, with the portion where the plane is higher counted negatively. The correction is to use the absolute height, or split the rectangle at the crossing and integrate $1-x$ on the left and $x-1$ on the right. That gives $1$, as shown in the worked example.

Key takeaway: Volume is always nonnegative. Integrate the upper height minus the lower height at each point; do not assume that $f-k$ is the correct height unless the surface is above the plane throughout the rectangle.

Why the Method Matters

The same setup applies whenever a solid is bounded below and above by two surfaces that are graphs over a rectangular base. For a horizontal plane, one vertical bound is constant and the other is $f(x,y)$, so the column height is the difference between them. Recognizing that height is the essential geometric step; the remaining work is evaluating a double integral using the rectangular bounds.

In a physical application, the coordinates may carry units. If $x$ and $y$ are measured in metres and $z$ is also measured in metres, the height has units of metres, the base area has units of square metres, and the volume has units of cubic metres. The integral’s units therefore provide a useful check that the calculation represents volume.

Check Your Understanding

For each question, identify the vertical height before evaluating or setting up an integral.

Over $R=[0,1]\times[0,2]$, the graph is $z=3+x$ and the plane is $z=3$. What height should be integrated?
If $f(x,y)\le k$ everywhere on a rectangle, which difference gives the nonnegative height?
Over $0\le x\le2$, what is wrong with using $\int_0^2(x-1)\,dx$ as the volume between $z=x$ and $z=1$?
Write an iterated integral for the volume between $z=5-x^2-y$ and $z=2$ over $[-1,1]\times[0,1]$, assuming the graph stays above the plane.
Why does the integral of a nonnegative height over the base have units of volume?